Difference between revisions of "1997 PMWC Problems/Problem I15"
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| − | 26 | + | == Problem == |
| + | How many paths from <math>A</math> to <math>B</math> consist of exactly six line segments (vertical, horizontal or inclined)? | ||
| + | |||
| + | <asy> | ||
| + | for(int i = 0; i < 3; ++i){ | ||
| + | draw((0,i+1)--(0,i)--(4,i)--(4,i+1)); | ||
| + | draw((4/3,i+1)--(4/3,i)--(8/3,i+1)--(8/3,i)); | ||
| + | } | ||
| + | draw((0,3)--(4,3)); | ||
| + | label("$A$",(0,0),SW); | ||
| + | label("$B$",(4,3),NE); | ||
| + | //Credit to chezbgone2 for the diagram</asy> | ||
| + | |||
| + | == Solution == | ||
| + | [[Casework]]: | ||
| + | |||
| + | *Ignoring the diagonal segments, there are <math>\frac{6!}{3!3!} = 20</math> paths. | ||
| + | *Traversing the diagonals, we quickly find that the path must run through exactly 2 diagonals. There are <math>{3\choose2} = 3</math> pairs of diagonals through which this is possible; quick counting shows us that each pair of diagonals yields 2 paths. So there are 6 more cases here. | ||
| + | |||
| + | In total, we get <math>20 + 6 = 26</math> paths. | ||
| + | |||
| + | == See Also == | ||
| + | {{PMWC box|year=1997|num-b=I14|num-a=T1}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Latest revision as of 14:37, 20 April 2014
Problem
How many paths from 
to 
consist of exactly six line segments (vertical, horizontal or inclined)?
![[asy] for(int i = 0; i < 3; ++i){ draw((0,i+1)--(0,i)--(4,i)--(4,i+1)); draw((4/3,i+1)--(4/3,i)--(8/3,i+1)--(8/3,i)); } draw((0,3)--(4,3)); label("$A$",(0,0),SW); label("$B$",(4,3),NE); //Credit to chezbgone2 for the diagram[/asy]](http://latex.artofproblemsolving.com/e/3/3/e33d97721111b81ed3553dceaad6fedd63bd16fe.png)
Solution
- Ignoring the diagonal segments, there are
paths. - Traversing the diagonals, we quickly find that the path must run through exactly 2 diagonals. There are
pairs of diagonals through which this is possible; quick counting shows us that each pair of diagonals yields 2 paths. So there are 6 more cases here.
paths.
pairs of diagonals through which this is possible; quick counting shows us that each pair of diagonals yields 2 paths. So there are 6 more cases here.In total, we get 
paths.
See Also
| 1997 PMWC (Problems) | ||
| Preceded by Problem I14 |
Followed by Problem T1 | |
| I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
